If your solution’s final volume of 32.55 mL yields 3.6 grams of AgCl, determine the Molarity of the final solution. Also, how many milliliters of a 1.85 M solution of AgNO3 did you add to an excess of
aqueous NaCl to provide the 3.6 grams of AgCl?
Actually this question is two parts.
The first part is to find the Final Molarity which is
(3.6 g * 1 mole /143.35 g) / (32.55 ml * 1 L /1000 ml) = 0.77 mol/L or M
The second part is to find how many millilitres of AgNo3 which has 1.85 M.
I don't understand how to do this second part.
Please help me this.
To solve the second part of the question, you need to use stoichiometry to determine the amount of AgNO3 required to produce the desired amount of AgCl.
First, you need to find the moles of AgCl produced. We already know from the first part of the question that it is 3.6 grams. To convert this to moles, divide by the molar mass of AgCl (143.35 g/mol):
moles of AgCl = 3.6 g / 143.35 g/mol = 0.0251 mol
Next, you need to determine the stoichiometric ratio between AgCl and AgNO3. From the balanced chemical equation, we know that 1 mole of AgCl reacts with 1 mole of AgNO3. Therefore, 0.0251 moles of AgCl will react with 0.0251 moles of AgNO3.
Now you can calculate the volume of the 1.85 M AgNO3 solution required. The molarity (M) of a solution is defined as the number of moles of solute per liter of solution. Rearranging the equation, you can calculate the volume (V) of solution required:
moles of AgNO3 = Molarity of AgNO3 * volume of AgNO3 (in liters)
0.0251 mol = 1.85 mol/L * V L
Solving for V:
V = 0.0251 mol / 1.85 mol/L = 0.0136 L = 13.6 mL
Therefore, you would need to add 13.6 mL of the 1.85 M AgNO3 solution to an excess of aqueous NaCl to produce 3.6 grams of AgCl.